First, we prove \(\rho _{A_h}\lesssim h^{-2}\) .
By the definition of the operator \(A_h\) , we have
\((A_h \widehat{u}_h, \widehat{u}_h)_{0,h}\simeq \Vert (\mathcal {L}_h(\widehat{u}_h),\widehat{u}_h)\Vert _{1,h}^2.\) 
Taking \(\mu =\widehat{u}_h\)  and \(v_h = \mathcal {L}_h(\widehat{u}_h)\) 
in (REF ) and applying the Cauchy-Schwarz inequality, we have
\(a_h((\mathcal {L}_h(\widehat{u}_h), 0), (\mathcal {L}_h(\widehat{u}_h),0)) = &\;-a_h((0, \widehat{u}_h), (\mathcal {L}_h(\widehat{u}_h),0))\\\le \;a_h((\mathcal {L}_h(\widehat{u}_h), 0), &(\mathcal {L}_h(\widehat{u}_h),0))^{1/2}a_h((0, \widehat{u}_h), (0,\widehat{u}_h))^{1/2}.\) 

Hence,
\(\Vert (\mathcal {L}_h(\widehat{u}_h),0)\Vert _{1,h}^2\simeq a_h((\mathcal {L}_h(\widehat{u}_h), 0), (\mathcal {L}_h(\widehat{u}_h),0)) \lesssim a_h((0, \widehat{u}_h), (0,\widehat{u}_h))\simeq \Vert (0,\widehat{u}_h)\Vert _{1,h}^2.\) 
Invoking the triangle inequality, we get
\((A_h \widehat{u}_h, \widehat{u}_h)_{0,h}\simeq \Vert (\mathcal {L}_h(\widehat{u}_h),\widehat{u}_h)\Vert _{1,h}^2\lesssim \Vert (0,\widehat{u}_h)\Vert _{1,h}^2= h^{-2} \Vert \widehat{u}_h\Vert _{0,h}^2,\) 
where the last equality following from the definition of these norms.
Since the spectral radius of the SPD operator \(A_h\)  is
its maximum eigenvalue,
\(\lambda _{A_h}^{\max }\) , we
conclude that
\(\rho _{A_h}=\lambda _{A_h}^{\max }\lesssim h^{-2}\) .
Next, we prove the condition number estimate.
By inverse and triangle inequalities, we get, for any \(\widehat{u}_h\in \widehat{W}_{h,0}^{k-1}\) ,
\(\min \lbrace \tau +1, h^{-2}\rbrace \Vert \widehat{u}_h\Vert _{0,h}^2\lesssim &\;(\tau +1) \Vert \mathcal {L}_h(\widehat{u}_h)\Vert _{0,h}^2+h^{-2} \Vert \mathcal {L}_h(\widehat{u}_h)-\widehat{u}_h\Vert _{0,h}^2\\\lesssim &\;(\tau +1) \Vert \mathcal {L}_h(\widehat{u}_h)\Vert _{\Omega }^2+h^{-2} \Vert \mathcal {L}_h(\widehat{u}_h)-\widehat{u}_h\Vert _{0,h}^2\\\lesssim &\;\Vert (\mathcal {L}_h(\widehat{u}_h), \widehat{u}_h)\Vert _{1,h}^2\simeq (A_h \widehat{u}_h, \widehat{u}_h)_{0,h},\) 

where in the last step, we invoked the
following discrete Poincaré inequality
\(\Vert \mathcal {L}_h(\widehat{u}_h)\Vert _\Omega \lesssim \Vert (\mathcal {L}_h(\widehat{u}_h), \widehat{u}_h)\Vert _{1,h},\) 
which holds on convex domains, see [1]}.
Hence, the minimum eigenvalue \(\lambda _{A_h}^{\min }\)  of \(A_h\) 
satisfies \(\min \lbrace \tau +1, h^{-2}\rbrace \lesssim \lambda _{A_h}^{\min }\) . This implies that
\(\kappa (A_h) = \frac{\lambda _{A_h}^{\max }}{\lambda _{A_h}^{\min }}\lesssim \frac{h^{-2}}{\min \lbrace \tau +1,\, h^{-2}\rbrace }.\) 
Finally, we prove \(h^{-2}\lesssim \rho _{A_h}\) .
Take \(\mu \)  as a basis function
of \(\widehat{W}_{h,0}^{k-1}\)  that is supported on a single facet \(F\) , then
norm equivalence of finite dimensional spaces and the
standard scaling argument implies that
\((A_h \mu , \mu )_{0,h}\simeq h^{-2}\Vert \mu \Vert _{0,h}^2.\) 

This concludes the proof.
